为什么在C#中浮动.NaN!= double.NaN？

Question

为什么float.NaN != double.NaN？

而float.PositiveInfinity == double.PositiveInfinity和float.NegativeInfinity == double.NegativeInfinity是相等的.

例:

bool PosInfinity = (float.PositiveInfinity == double.PositiveInfinity); //true
bool NegInfinity = (float.NegativeInfinity == double.NegativeInfinity); //true

bool isNanEqual = (float.NaN == double.NaN);  //false, WHY?

Answer 1

NaN永远不会等于NaN(甚至在同一类型内).因此存在IsNaN函数的原因:

Double zero = 0;
// This will return true.
if (Double.IsNaN(0 / zero)) 
{
    Console.WriteLine("Double.IsNan() can determine whether a value is not-a-number.");
}

您还应该知道,您所显示的所有比较都不是"按原样"发生的.在编写时floatValue == doubleValue,浮点数实际上会在比较发生之前隐式转换为双精度数.

Answer 2

可能是因为 NaN != NaN

Answer 3

引用维基百科:

与NaN进行比较时,即使与自身进行比较,也会返回无序结果.